A is a solution of dibasic acid H2X. B is a solution of 1.00 g sodium hydroxide in 250 cm³ of solution. Put A into the burette and titrate against 20.00 cm³ or 25.00 cm³ portion of B using methyl orange as an indicator. Repeat the titration three more times and record your results in Table 1.1

The instructions state that a "20.00 cm$^3$ or 25.00 cm$^3$ portion of B" should be used. You must choose one of these standard pipette volumes for your experiment and record it. For example, if you choose a 25.00 cm$^3$ pipette, you would write:
Volume of pipette used: $\boxed{\text{25.00 cm}^3}$ (or 20.00 cm$^3$ if that was chosen)

Given:
Mass of NaOH = $1.00 \text{ g}$
Volume of solution = $250 \text{ cm}^3$

First, calculate the molar mass of NaOH:
Molar mass of NaOH $= (22.99 \text{ g/mol for Na}) + (16.00 \text{ g/mol for O}) + (1.01 \text{ g/mol for H})$
Molar mass of NaOH $= 40.00 \text{ g/mol}$

Next, calculate the number of moles of NaOH:
Moles of NaOH $= \frac{\text{Mass}}{\text{Molar mass}}$
Moles of NaOH $= \frac{1.00 \text{ g}}{40.00 \text{ g/mol}}$
Moles of NaOH $= 0.0250 \text{ mol}$

Convert the volume of solution to dm$^3$:
Volume $= 250 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3}$
Volume $= 0.250 \text{ dm}^3$

Finally, calculate the concentration of NaOH:
Concentration of NaOH $= \frac{\text{Moles}}{\text{Volume}}$
Concentration of NaOH $= \frac{0.0250 \text{ mol}}{0.250 \text{ dm}^3}$
Concentration of NaOH $= \boxed{0.100 \text{ mol/dm}^3}$

For each titration (Rough, 1, 2, 3), you will record the initial and final burette readings. The "Volume of A used" is then calculated for each titration using the formula:

The rough titration is an estimate, while titrations 1, 2, and 3 are precise measurements. You would typically use the concordant (closest) values from titrations 1, 2, and 3 to calculate an average volume of A used for subsequent calculations.